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3 years ago

Major species present when fructose is dissolved in water

Chemistry

2 answers:

Katyanochek1 [597]3 years ago

7 0

The fructose chemical formula is C6H12O6. The answer to the question above regarding the major species present when fructose is dissolved in water (H2O) is "None". No ions are present. It is false that when sugar is dissolved in water there will be strong electrolytes.

BARSIC [14]3 years ago

6 0

Answer:

Alfa-fructofuranose and beta-fructofuranose

Explanation:

The species more stables in water of fructose are their cyclic forms. In aqueous solution, the ketone group of the fructose reacts in a reversible way with the hydroxyl group of the same forming a cyclic hemiketal, in a reaction of intramolecular cycling. These rings formed are more stables when contain 5 or 6 atoms, named furanose and pyranose forms, respectively. In this cycling process the carbonyl carbon (C=O) it is transformed into a chiral center which produces 2 possible anomers, alfa and beta. In the fructose, the hydroxyl group of the formed hemiketal it is produced in the carbon 2 (the anomeric carbon) and can be located above the ring or beneath the ring. In D-sugars (the most common in nature), when the hydroxyl group is located beneath the ring the structure is in the alfa form and when is above the ring is in the beta form.

In the case of fructose dissolved in water it is formed the alfa-fructofuranose an beta-fructofuranose in a majority way, as shown in the picture.

Sugars are capable of suffering of mutarotation when they are dissolved in water, which cause the interconversion of anomers alfa and beta, producing a mix of the alfa and beta forms in the furanose or pyranose rings. And the proportion of each

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2 years ago

Why might increasing the temperature alter the rate of a chemical reaction?

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3 years ago

What mass of lithium phosphate would you mass to make 2.5 liter of 1.06 M lithium

alex41 [277]

Answer:

Approximately $3.06 \times 10^{2}\; \rm g$ (approximately $306\; \rm g$ .)

Explanation:

Calculate the quantity $n$ of lithium phosphate in $V = 2.5\; \rm L$ of this $c = 1.06\; \rm M = 1.06\; \rm mol \cdot L^{-1}$ lithium phosphate solution.

$\begin{aligned}n &= c \cdot V\\ &= 2.5\; \rm L \times 1.06\; mol \cdot L^{-1}\\ &= 2.65\; \rm mol\end{aligned}$ .

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$\rm Li$ : $6.94$ .
$\rm P$ : $30.974$ .
$\rm O$ : $15.999$ .

Calculate the formula mass of $\rm Li_3PO_4$ :

$M(\rm Li_3PO_4) = 3 \times 6.94 + 30.974 + 4 \times 15.999 = 115.79\; \rm g \cdot mol^{-1}$ .

Calculate the mass of that $n = 2.65\; \rm mol$ of $\rm Li_3PO_4$ formula units:

$\begin{aligned}m &= n \cdot M \\ &= 2.65\; \rm mol \times 115.79\; \rm g\cdot mol^{-1} \\ &\approx 3.06 \times 10^{2}\; \rm g \end{aligned}$ .

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